**Find Coordinates of Centroid of Triangle :**

Here we are going to see some example problems on finding centroid of triangle.

The centroid G of the triangle with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) is

G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ]

**Question 1 :**

Find the centroid of the triangle whose vertices are

(i) (2, −4), (−3, −7) and (7, 2)

**Solution :**

Let the points be A (2, −4) B (−3, −7) and C (7, 2)

= G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ]

= (2 - 3 + 7)/3, (-4 - 7 + 2)/3

= (6/3, -9/3)

= (2, -3)

(ii) (−5,−5), (1,−4) and (−4,−2)

**Solution :**

Let the points be A (−5,−5) B (1,−4) and C (-4, -2)

= G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ]

= (-5 + 1 - 4)/3, (-5 - 4 - 2)/3

= (-9 + 1)/3, (-11/3)

= (-8/3, -11/3)

**Question 2 :**

If the centroid of a triangle is at (4,−2) and two of its vertices are (3,−2) and (5,2) then find the third vertex of the triangle

**Solution :**

Let the given vertices be A(3, -2) and B (5, 2)

Centroid of the triangle = (4, -2)

Let the third vertex be C (a, b)

= G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ]

(3 + 5 + a)/3, (-2 + 2 + b)/3 = (4, -2)

(8 + a)/3, (b/3) = (4, -2)

By equating x and y-coordinates, we get

(8 + a)/3 = 4 8 + a = 12 a = 12 - 8 a = 4 |
b/3 = -2 b = -6 |

Hence the required vertex is (4, -6).

**Question 3 :**

Find the length of median through A of a triangle whose vertices are A(−1, 3), B(1, −1) and C(5, 1).

**Solution :**

If we draw a median through A, it will intersect the side BC exactly at middle.

Mid point of the side BC = D

= (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

= (1 + 5)/2, (-1 + 1)/2

= (6/2, 0/2)

= (3, 0)

Length of median AD = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

= √(3 + 1)^{2} + (0 - 3)^{2}

= √4^{2} + (-3)^{2}

= √(16 + 9)

= √25

= 5 units.

After having gone through the stuff given above, we hope that the students would have understood, "Find Coordinates of Centroid of Triangle"

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